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I'm tring to prove that the unit circle $$S^1=\{(x_1,x_2)\in\mathbb{R}^2\text{ such that }x_1^2+x_2^2=1\}$$ is an embedded submanifold of $\mathbb{R}^2$ using the following Characterization:

A nonempty subset $M \subset\mathbb{R}^n$ is an m-manifold iff:

For every $p\in M$, there are two open sets $O,W\subset\mathbb{R}^n$ with $0_n\in O$ and $p ∈ M ∩ W$, and a smooth diffeomorphism $ϕ: O → W$, such that $ϕ(0_n) = p$ and $$ϕ(O ∩ (\mathbb{R}^m × {0_{n−m}})) = M ∩ W$$.

My attempt:

lets fix $a\in S^1$, there exists a unique $\theta_0\in[0,2\pi)$ such that $a=(cos(\theta_0),sin(\theta_0))$

and let $\phi$ be the diffeomorphism \begin{array}{cccc} \phi : & (0,\infty)\times(\theta_0-\pi,\theta_0+\pi) & \longrightarrow & \mathbb{R}^2\backslash D_{\theta_0+\pi}\\ ~~ & (r,\theta) & \mapsto &(rcos(\theta),rsin(\theta)) \end{array}

where $D_{\theta_0+\pi}$ is the half line at the origin with polar angle $\theta_0$

From here i dont know how to proceed. Should i modify this map or use as it is? and who are the two open sets of the charactrization?

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The way you constructed your $\phi$ is almost good. Intuitively, the characterization you presented requires $\phi$ to be a diffeomorphism between 2 open sets $W$ - the ambient space around $\mathbb S^1$, and $O$ - a deformed version of $W$ such that $\mathbb S^1 \cap W$ is "squished and flattened" into $\mathbb R \times \left \{ 0 \right \}$ (by $\phi^{-1}$). You could think of $\phi|_{O\cap (\mathbb R \times \left \{ 0 \right \})}$ as a parameterization of the circle using $\theta$.

Your $\phi$ is a good candidate besides the fact there needs to be some alignment so that $0$ is mapped to $a$: substituting $\theta+\theta_0$ into the trig functions instead of $\theta$, and using a radius $1+r$ instead of $r$. This way we get $W=\mathbb R^2 \setminus D_{\theta_0+\pi}$ as you said but $$O=(-1,\infty)\times(-\pi,\pi)$$ From the naturality of polar coordinates it's pretty easy to convince $\phi$ is bijective, two points can't have the same radius and angle, and every point has a radius and an angle. $\phi$ is also smooth due to its components being elementary functions. Hence, it is a diffeomorphism.

$\phi(O\cap (\mathbb R \times \left \{ 0 \right \}))\subseteq \mathbb S^1 \cap W$ since any point of the form $(0,\theta)$ is mapped to the unit circle at an angle which isn't $\theta_0+\pi$, and $\mathbb S^1 \cap W \subseteq\phi(O\cap (\mathbb R \times \left \{ 0 \right \})) $ using the reverse argument.

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  • $\begingroup$ thank you sir you're a genius So we chose our diffeomorphism to be \begin{array}{cccc} \phi : & (-1,\infty)\times(-\pi,+\pi) & \longrightarrow & \mathbb{R}^2\backslash D_{\theta_0+\pi}\\ ~~ & (r,\theta) & \mapsto &((1+r)cos(\theta+\theta_0),(1+r)sin(\theta+\theta_0)) \end{array} $O=(-1,\infty)\times(-\pi,+\pi)$ and $w=\mathbb{R}^2\backslash D_{\theta_0+\pi}$ $\endgroup$ – Donnie Darko Oct 23 '20 at 18:28
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    $\begingroup$ Yup, I think that should work @DonnieDarko $\endgroup$ – Theorem Oct 23 '20 at 19:02

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